For a vector $v \in [0,1]^n$ and $p > 1$ we denote the p-norm of $v$ as: $||v||_p = (\sum_iv_i^p)^{\frac{1}{p}}$. where $v_i$ are the entries of $v$.
Define the following (weird looking) function $f:[0,1]^n \to \mathbb{R}$. $f(v) = \frac{||v||_2^{10}}{||v||_4^8}\exp(-\frac{||v||_2^4}{||v||_4^4})$.
I'm looking for ways to bound this function. An easy bound to see using Holder's inequality is that $f(v) \leq \frac{n}{e}$. The question is, can we do better?
I believe there should be a bound which is independent of the dimension. But I can't seem to find it (or show the converse). Any help would be much appreciated.
You are correct --- There is indeed a bound independent of dimension.
Notice that $f(tv)=t^2 f(v)$ for $t>0$ due to the homogeneity of the norms. By consequence, $f(v)$ will be maximized when $v\in[0,1]^n$ satisfies $\max_i v_i = 1$. So suppose $v$ has this property.
Because $0\le v_i\le 1$ for each $i$ and $v_i=1$ for some $i$ we have $$ 1\le \|v\|_4^4 = \sum_i v_i^4 \le \sum_i v_i^2 = \|v\|_2^2. $$ Therefore $\|v\|_2^2/\|v\|_4^4 \ge 1$, and it follows $$ f(v) = \frac{\|v\|_2^{10}}{\|v\|_4^8} \exp(- \frac{\|v\|_2^2}{\|v\|_4^4}\|v\|_2^2) \le \frac{\|v\|_2^{10}}1 \exp(-\|v\|_2^2) \le \max_{x>0} x^5 \exp(-x) = \left(\frac 5e\right)^5. $$